A) 34
B) 33
C) 66
D) 68
Correct Answer: A
Solution :
\[{{a}_{1}}+{{a}_{5}}+.......+{{a}_{49}}=416\] \[{{a}_{1}}+{{a}_{1}}+4d+.......+{{a}_{1}}+48d=416\] \[13{{a}_{1}}+4d\left( \frac{12\times 13}{2} \right)=416\] \[13({{a}_{1}}+24d)=416\] \[\therefore {{a}_{1}}+24d=32\]??(1) Also, \[{{a}_{9}}+{{a}_{43}}=66\] \[\therefore 2{{a}_{1}}+50d=66\] \[{{a}_{1}}+25d=33\] ?..(2) From (1) & (2), we get \[d=1\] & \[{{a}_{1}}=8\] \[\therefore \] \[{{a}_{1}}^{2}+{{a}_{2}}^{2}+......+{{a}_{17}}^{2}\] \[={{8}^{2}}+{{9}^{2}}+......+{{24}^{2}}\] \[=\Sigma {{24}^{2}}-\Sigma {{7}^{2}}\] \[=140\times 34\]You need to login to perform this action.
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